已知
\lim_{x\to 0}\frac{\ln(1-x) + xf(x)}{x^2} = 0
求
\lim_{x\to 0} \frac{f(x) - 1}{x}

由题意，可知
\frac{\ln(1-x) + xf(x)}{x^2} = o(x^2)

故

f(x) = \frac{o(x^2) - \ln(1-x)}{x}

故

\lim_{x\to 0} \frac{f(x) - 1}{x} = \lim_{x\to 0} \frac{o(x^2) - \ln(1-x) - x}{x^2} \\ = \lim_{x\to 0} \frac{- [(-x) - \frac{(-x)^2}{2} + o(x^2)] - x}{x^2} = \lim_{x\to 0} \frac{-\frac{x^2}{2}}{x^2} = -\frac{1}{2}